Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

Why isn't g(x)= 0 why does it have to be f(x)? Why not both?

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- Jul 8th 2011, 10:35 AMCountingPenguinsReimann integral question
Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

Why isn't g(x)= 0 why does it have to be f(x)? Why not both? - Jul 8th 2011, 10:53 AMgirdavRe: Reimann integral question
What we suppose works for all Riemann-integrable function $\displaystyle g$, and we only know that $\displaystyle \int_a^b fg =0$.

- Jul 8th 2011, 11:09 AMPlatoRe: Reimann integral question
- Jul 8th 2011, 12:29 PMCountingPenguinsRe: Reimann integral question
f is continuous and g is integrable and the product of the integral of fg over the interval [a,b] is 0. So f(x) must be 0 for all x in [a,b], is all the information I have been given. I will start with the definition of continuous for f and integrable for g and then hope I see something useful after that.

- Jul 8th 2011, 12:45 PMPlatoRe: Reimann integral question
Did you see reply #3. That answers the question.

- Jul 10th 2011, 01:49 PMAlso sprach ZarathustraRe: Reimann integral question
I came up with the following solution, I hope it is not so wrong. (Itwasntme)

Suppose that $\displaystyle f$ is not $\displaystyle 0$ for all $\displaystyle x\in[a,b]$, then there exists $\displaystyle x_0\in[a,b]$ such that $\displaystyle f(x_0)>0$ (If the is no such $\displaystyle x_0$, there must be $\displaystyle x_1\in[a,b]$ such that $\displaystyle f(x_1)<0$.In that case we will look at $\displaystyle -f$).

f is continuous function therefor exists $\displaystyle \delta>0$ for which $\displaystyle f(x)>\epsilon>0$ for all $\displaystyle x\in (x_0-\delta,x_0+\delta)$.

Now we define $\displaystyle h(x)$-'staircase'-characteristic function of open interval $\displaystyle (x_0-\delta,x_0+\delta)$.

$\displaystyle h(x)=\left\{\begin{matrix}1 & x\in (x_0-\delta,x_0+\delta) \\ 0 & \text{otherwise} \end{matrix}\righ$

Hence, we will get:

$\displaystyle 0=\int_{a}^{b}f(x)h(x)dx=\int_{a}^{x_0-\delta}f(x)h(x)dx+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+\int_{x_0+\delta}^{ b}f(x)h(x)dx=0+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+0>2\epsilon\delta $.

So, we got a contradiction to the fact that if $\displaystyle f$ is continuous on $\displaystyle [a,b]$ and h is staircase function $\displaystyle \int_{a}^{b}f(x)h(x)dx=0 \Rightarrow f\equiv 0$.

But we also know that if $\displaystyle g$ is integrable function on $\displaystyle [a,b]$, so for every $\displaystyle \epsilon_0>0$ exists staircase function $\displaystyle h$ so that

$\displaystyle \int_{a}^{b}|g(x)-h(x)|<\epsilon_0$

Combining these two results we will conclude the needed. - Jul 10th 2011, 03:35 PMPlatoRe: Reimann integral question
Your solution is correct.

Note that $\displaystyle f \text{ is continuous on }[a,b]$ so as such that means that both $\displaystyle f~\&~f^2$ is are integrable on $\displaystyle [a,b]$.

By the given $\displaystyle \int_a^b {f \cdot f} = \int_a^b {f^2 } = 0$.

But by your reasoning that must imply that $\displaystyle f \equiv 0$.